1

Property Relationships

2

dUWQ revrev =−δδ

TdSQrev =δ PdVWrev =δ

dUPdVTdS =−

Apply the differential form of the first law for a closed stationary system for an reversible process

The T-ds relations:

3

PdvduTds +=This equation is known as:

First Gibbs equation or First Tds relationship

Divide by T, ..

TPdv

Tduds +=

Although we get this form for reversible process, we still can compute ∆s for an irreversible process. This is because S is a point function.

Divide by the mass, you get

4

Second T-ds (Gibbs) relationship Recall that… Pvuh +=

vdPPdvdudh ++=Take the differential for both sides

Rearrange to find du

Substitute in the First Tds relationship PdvduTds +=

vdPPdvdhdu −−=

vdPdhTds −=Second Tds relationship, or Gibbs equation

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TvdP

Tdhds −=Divide by T, ..

Thus We have two equations for ds

TvdP

Tdhds −=

TPdv

Tduds +=

To find ∆s, we have to integrate these equations. Thus we need a relation between du and T (or dh and T).

Now we can find entropy change (the LHS of the entropy balance) for liquids and solids

6

2- Entropy Change of Liquids and Solids

TPdv

Tduds +=

Solids and liquids do not change specific volume appreciably with pressure. That means that dv=0, so the first equation is the easiest to use.

0

Tduds =

Thus For solids and liquids

Recall also that For solids and liquids, CdTdu =

TCdT

Tduds ==∴

7

=∆

1

2lnTTCsIntegrate

to give…

Only true for solids and liquids!!

What if the process is isentropic? 0ln,0

1

2 =

⇒=∆

TTCs

The only way this expression can equal 0 is if,

Hence, for solids and liquids, isentropic processes are also isothermal.

21 TT =

8

3- The Entropy Change of Ideal Gases, first relation

The entropy change of an ideal gas can be obtained by substituting du = CvdT and P /T= R/υ into Tds relations:

( )1

2

2

1

12 υυlnR

TdTTCss v∫ +=−⇒

Tds du pdυ= +v

dT dds C RT

υυ

= +

integrating

First relation

du PddsT T

υ= +

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A second relation for the entropy change of an ideal gas for a process can be obtained by substituting dh = CpdT and υ/T= R/P into Tds relations:

( )∫ −=−2

1 1

212 P

PlnRTdTTCss p

Tds dh vdp= − pdT dpds C RT p

= −

integrating

Second relation

dh vdpdsT T

= −

10

( )1

22

112 ln

υυR

TdTTCss v∫ +=−

( )∫ −=−2

1 1

212 P

PlnRTdTTCss p

The integration of the first term on the RHS can be done via two methods:

1. Assume constant Cp and constant Cv (Approximate Analysis)

2. Evaluate these integrals exactly and tabulate the data (Exact Analysis)

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Method 1: Constant specific heats (Approximate Analysis) First relation

+

=∆

1

2

1

2 lnlnvvR

TTCs v

Only true for ideal gases, assuming constant heat capacities

Second relation

Only true for ideal gases, assuming constant heat capacities

−

=∆

1

2

1

2 lnlnPPR

TTCs p

( ) ⇒+=− ∫1

22

112 ln

υυR

TdTTCss v

( ) ⇒−=− ∫2

1 1

212 ln

PPR

TdTTCss p

12

Method 2: Variable specific heats (Exact Analysis)

We could substitute in the equations for Cv and Cp, and perform the integrations Cp = a + bT + cT2 + dT3 But this is time consuming.

1

22

1 PPlnR

TdTC

s p −=∆ ∫We use the second relation

7.5. Isentropic Processes of an ideal gas

13

14

Isentropic Processes of Ideal Gases Many real processes can be modeled as

isentropic

Isentropic processes are the standard against which we should measure efficiency

We need to develop isentropic relationships for ideal gases, just like we developed for solids and liquids

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+

=∆

1

2

1

2 lnlnvvR

TTCs v

For the isentropic case, ∆S=0. Thus

−=

1

2

1

2 lnlnvvR

TTCv

Constant specific heats (1st relation)

vCR

v vv

vv

CR

TT

=

−=

2

1

1

2

1

2 lnlnln

Recall

Recall also from previous chapter, the following relations..…

11 −=−=⇒−= kC/CC/RCCR vpvvp

1

2

1

1

2

−

=

∴

k

vv

TT Only applies to

ideal gases, with constant specific heats

16

0lnln1

2

1

2 =

−

=∆

PPR

TTCs p

pCR

p PP

PP

CR

TT

=

=

1

2

1

2

1

2 lnlnln

kk

PP

TT

1

1

2

1

2

−

=

∴

Only applies to ideal gases, with constant specific heats

Constant specific heats (2nd relation)

Recall..… or 1/ −= kCR v1/ p

kR Ck−

=

17

Since…

kk

PP

TT

1

1

2

1

2

−

=

1

2

1

1

2

−

=

k

vv

TT

and

kkk

PP

vv

1

1

2

1

2

1

−−

=

Which can be simplified to…

=

1

2

2

1

PP

vv

kThird isentropic relationship

HENCE